CBSE Class 10 Maths Advanced Real Numbers Short Answer Questions Set 01

Short Answer Questions

Question 1. Prove that \( \sqrt3+\sqrt7 \) is irrational.
Answer:
Suppose \( \sqrt3+\sqrt7 \) is a rational number r (r ≠ 0, since the sum is positive).
Then \( \sqrt7 = r-\sqrt3 \). Squaring: \( 7 = r^2-2r\sqrt3+3 \), which gives \( \sqrt3 = \frac{r^2-4}{2r} \).
The right side is rational, so \( \sqrt{3} \) would be rational. But \( \sqrt{3} \) is irrational, a contradiction.
Hence \( \sqrt3+\sqrt7 \) is irrational.
In simple words: If the sum were rational, we could isolate √3 as a fraction, which is impossible.

Teacher's Note:
1. Isolate one root, square, and the other root appears alone.
2. Use the known result that √3 is irrational.

Question 2. Explain why 3 × 5 × 7 + 7 is a composite number.
Answer:
3 × 5 × 7 + 7 = 7(3 × 5 + 1) = 7 × 16 = 112.
The number can be written as a product of two numbers each greater than 1 (7 and 16), so it has more than two factors.
Hence it is composite.
In simple words: Taking 7 as a common factor shows the number is 7 × 16.

Teacher's Note:
1. A number with a factor other than 1 and itself is composite.
2. The same trick works for 7 × 6 × 5 × ... + 5.

Question 3. Without actually performing the long division, find if \( \frac{987}{10500} \) will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.
Answer:
987 = 3 × 7 × 47 and 10500 = 2² × 3 × 5³ × 7.
Cancelling the common factors 3 and 7: \( \frac{987}{10500} = \frac{47}{500} \) and 500 = 2² × 5³.
In lowest terms the denominator has only the primes 2 and 5, so the decimal expansion is terminating (it equals 0.094).
In simple words: After cancelling common factors, only 2s and 5s remain in the denominator.

Teacher's Note:
1. Do not judge by the original denominator 10500; reduce first.
2. The prime 7 in 10500 cancels with the 7 in 987.

Question 4. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form \( \frac{p}{q} \) ?
Answer:
The decimal terminates after 4 places, so \( 327.7081 = \frac{3277081}{10^4} \).
The numerator is not divisible by 2 or 5, so the fraction is already in lowest terms, with q = 10⁴ = 2⁴ × 5⁴.
So q is of the form \( 2^m\times5^n \): its only prime factors are 2 and 5 (here m = n = 4).
In simple words: A terminating decimal always comes from a denominator that has only 2 and 5 as prime factors.

Teacher's Note:
1. Four decimal places mean the denominator is 10⁴.
2. Check that p is not divisible by 2 or 5 to be sure the fraction is in lowest terms.

Question 5. Explain 5 × 4 × 3 × 2 × 1 + 3 is a composite number. [CBSE 10th Board 2013, 2 Marks]
Answer:
5 × 4 × 3 × 2 × 1 + 3 = 120 + 3 = 123.
Taking 3 as a common factor: 3(5 × 4 × 2 × 1 + 1) = 3 × 41.
The number has factors other than 1 and itself (3 and 41), so it is composite.
In simple words: Both terms are multiples of 3, so the sum is also a multiple of 3.

Teacher's Note:
1. Look for a common factor in the two terms.
2. 123 = 3 × 41.

Question 6. Use Euclid's Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer:
Any positive integer n is 3q, 3q + 1 or 3q + 2.
n = 3q: n² = 9q² = 3(3q²) = 3m.
n = 3q + 1: n² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1.
n = 3q + 2: n² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1.
So a square is always of the form 3m or 3m + 1.
In simple words: Check the three possible remainders on division by 3 and square each.

Teacher's Note:
1. A square never leaves remainder 2 when divided by 3.
2. Check: 5² = 25 = 3 × 8 + 1.

Question 7. Show that of the numbers n, n + 2 and n + 4, only one of them is divisible by 3.
Answer:
Let n = 3q + r with r = 0, 1 or 2.
r = 0: n = 3q is divisible; n + 2 = 3q + 2 and n + 4 = 3(q + 1) + 1 are not.
r = 1: n + 2 = 3(q + 1) is divisible; n = 3q + 1 and n + 4 = 3(q + 1) + 2 are not.
r = 2: n + 4 = 3(q + 2) is divisible; n = 3q + 2 and n + 2 = 3(q + 1) + 1 are not.
So in every case exactly one of the three is a multiple of 3.
In simple words: The remainders of n, n + 2, n + 4 on division by 3 are r, r + 2, r + 1 (mod 3), all different, so one of them is 0.

Teacher's Note:
1. The three numbers leave three different remainders.
2. Try n = 5: the numbers 5, 7, 9; only 9 is divisible by 3.

Question 8. Let n = 640640640643, without actually computing n² prove that n² leave remainder 1 when divided by 8.
Answer:
n ends in 3, so n is odd. Write n = 2k + 1.
n² = 4k² + 4k + 1 = 4k(k + 1) + 1.
k and k + 1 are consecutive integers, so one of them is even and k(k + 1) is even, say k(k + 1) = 2t.
Then n² = 8t + 1, so n² leaves remainder 1 when divided by 8.
In simple words: Every odd square is one more than a multiple of 8.

Teacher's Note:
1. Check with 3² = 9 = 8 + 1 and 5² = 25 = 24 + 1.
2. The key fact is that k(k + 1) is always even.

Question 9. If d is the HCF of 56 and 72, find x, y satisfying d = 56x + 72y. Also, show that x and y are not unique.
Answer:
72 = 56 × 1 + 16
56 = 16 × 3 + 8
16 = 8 × 2 + 0
So d = 8.
Back-substitute: 8 = 56 − 3 × 16 = 56 − 3(72 − 56) = 4 × 56 − 3 × 72. So one solution is x = 4, y = −3.
To get another, add and subtract 56 × 72: 8 = 56 × 4 + 72 × (−3) + 56 × 72 − 72 × 56 = 56 × (4 + 72) + 72 × (−3 − 56) = 56 × 76 + 72 × (−59).
So x = 76, y = −59 also works (check: 4256 − 4248 = 8). Two different pairs satisfy the relation, so x and y are not unique.
In simple words: Write the HCF as a combination of 56 and 72; adding and subtracting 56 × 72 gives a different pair.

Teacher's Note:
1. Back-substitution through the Euclid steps gives the first pair.
2. Infinitely many pairs exist: (4 + 9t, −3 − 7t).

Question 10. Using Euclid's Division algorithm, find the greatest number that divides 445, 572 and 699 leaving remainder 4, 5 and 6 respectively.
Answer:
Subtract the remainders: 445 − 4 = 441, 572 − 5 = 567, 699 − 6 = 693. The required number is the H.C.F. of 441, 567 and 693.
H.C.F.(567, 441):
567 = 441 × 1 + 126
441 = 126 × 3 + 63
126 = 63 × 2 + 0
So H.C.F.(441, 567) = 63. Now check 693: 693 = 63 × 11 + 0, so 63 divides 693.
The greatest number is 63 (it is larger than the remainders 4, 5, 6).
In simple words: Take the HCF of the numbers after deducting their remainders.

Teacher's Note:
1. Do the algorithm for two numbers first, then test the third.
2. The divisor must exceed the largest remainder, 6.

Question 11. Find the HCF and LCM of the following using Fundamental Theorem of Arithmetic method.
(i) 426 and 576   (ii) 625, 1125 and 2125
Answer:
(i) 426 = 2 × 3 × 71 and 576 = 2⁶ × 3². Common primes 2 and 3 with smallest powers: H.C.F. = 2 × 3 = 6. L.C.M. = 2⁶ × 3² × 71 = 64 × 9 × 71 = 40896.
(ii) 625 = 5⁴, 1125 = 3² × 5³, 2125 = 5³ × 17. Only 5 is common to all, with smallest power 5³: H.C.F. = 125. L.C.M. = 5⁴ × 3² × 17 = 625 × 153 = 95625.
In simple words: Factorise into primes, then use smallest common powers for the HCF and biggest powers for the LCM.

Teacher's Note:
1. A prime missing from even one number is left out of the HCF.
2. Check (i): 6 × 40896 = 245376 = 426 × 576.

Question 12. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
First number: 7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 1014. It is a product of two numbers bigger than 1, so it is composite.
Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045 = 5(1008 + 1) = 5 × 1009. Again it has factors other than 1 and itself, so it is composite.
In simple words: Both terms share a common factor, which can be taken out to show a non-trivial factorisation.

Teacher's Note:
1. 13 is common in the first sum and 5 in the second.
2. Any number with a factor other than 1 and itself is composite.

Question 13. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
Answer:
The number of columns must divide both 616 and 32, and we want the largest, so it is the H.C.F.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
H.C.F. = 8, so the maximum number of columns is 8.
In simple words: Same number of columns for both groups means a common divisor; the maximum is the HCF.

Teacher's Note:
1. Translate the story into 'greatest common divisor'.
2. Check: 616 ÷ 8 = 77 and 32 ÷ 8 = 4 members per column.

Question 14. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?
Answer:
They meet at the start again after a time that is a multiple of both 18 and 12, and the first such time is the L.C.M.
18 = 2 × 3² and 12 = 2² × 3, so L.C.M. = 2² × 3² = 36.
They meet again at the starting point after 36 minutes.
In simple words: The first common multiple of their round times is the LCM, 36 minutes.

Teacher's Note:
1. Meeting again at the same point means whole numbers of rounds for both.
2. Sonia does 2 rounds and Ravi does 3 rounds in 36 minutes.

Question 15. Aakash, Kushal and Harish go for a morning walk. They step off together and their steps measure 40cm, 42cm and 45cm, respectively. (a) What is the minimum distance each should walk so that each can cover the same distance in complete steps? (b) How is morning walk useful?
Answer:
(a) The distance must be a multiple of 40, 42 and 45, so the minimum is their L.C.M.
40 = 2³ × 5, 42 = 2 × 3 × 7, 45 = 3² × 5. L.C.M. = 2³ × 3² × 5 × 7 = 2520 cm (25 m 20 cm).
(b) A morning walk keeps the body fit, improves the heart and lungs, helps control weight and refreshes the mind for the day, so it is good for health.
In simple words: The shortest common distance is the LCM of the step lengths; walking regularly is good for health.

Teacher's Note:
1. Convert the final answer to metres if needed: 2520 cm = 25.2 m.
2. For the value part, give a clear health reason.

Question 16. Prove that \( \sqrt{3} \) is an irrational number.
Answer:
Suppose, to the contrary, that \( \sqrt{3} \) is rational. Then \( \sqrt{3} = \frac{a}{b} \) where a and b are co-prime positive integers.
Squaring: \( a^2 = 3b^2 \), so 3 divides a². Since 3 is prime, it divides a. Write a = 3c.
Then \( 3^2c^2 = 3b^2 \), so \( b^2 = 3c^2 \). Now 3 divides b², hence divides b.
So 3 is a common factor of a and b, contradicting that they are co-prime. Therefore \( \sqrt{3} \) is irrational.
In simple words: Assuming √3 is a fraction forces both numerator and denominator to be multiples of 3, which is impossible.

Teacher's Note:
1. Use the fact that if a prime divides a², it divides a.
2. The same method works for √2, √5, √7 and any prime.

Question 17. Prove that \( 5-2\sqrt3 \) is an irrational number.
Answer:
Suppose \( 5-2\sqrt3 \) is rational, equal to r.
Then \( 2\sqrt3 = 5-r \), so \( \sqrt3 = \frac{5-r}{2} \), which is rational (a difference of rationals divided by 2).
This contradicts the fact that \( \sqrt{3} \) is irrational. Hence \( 5-2\sqrt3 \) is irrational.
In simple words: If 5 − 2√3 were rational, √3 could be written as a fraction.

Teacher's Note:
1. Isolate the surd on one side.
2. Rational ± (rational × irrational) stays irrational.